A tightly-wound solenoid of radius a and length l has n turns per...

Let the center of the solenoid be the origin. The distance of the elemental ring from the center is $y$.

The magnetic field at the center due to an elemental ring of width $dy$ at a distance $y$ from the center is:

$dB = \dfrac{\mu_0 i (n \, dy) a^2}{2(a^2 + y^2)^{3/2}}$

Integrating from $y = -\dfrac{l}{2}$ to $y = \dfrac{l}{2}$ to find the total magnetic field:

$B = \int_{-l/2}^{l/2} \dfrac{\mu_0 n i a^2}{2(a^2 + y^2)^{3/2}} dy$

Let $y = a \tan\theta$, then $dy = a \sec^2\theta \, d\theta$.

$B = \dfrac{\mu_0 n i a^2}{2} \int \dfrac{a \sec^2\theta}{(a^2 \sec^2\theta)^{3/2}} d\theta = \dfrac{\mu_0 n i}{2} \int \cos\theta \, d\theta$

$B = \dfrac{\mu_0 n i}{2} [\sin\theta]$

Substituting $\sin\theta = \dfrac{y}{\sqrt{a^2 + y^2}} and applying the limits from$y = -\dfrac{l}{2}$to$y = \dfrac{l}{2}$:<br><br>$B = \dfrac{\mu_0 n i}{2} \left[\dfrac{l/2}{\sqrt{a^2 + (l/2)^2}} - \dfrac{-l/2}{\sqrt{a^2 + (-l/2)^2}} \right]

$B =$\dfrac{\mu_0 n i}{2} \left( \dfrac{l}{\sqrt{a^2 + l^2/4}} \right) = \dfrac{\mu_0 n i}{\sqrt{1 + \left(\dfrac{2a}{l}\right)^2}}